Question

a plane left 30 minutes late then its scheduled time and in order to reach the destination 1500 km away in X it had to increase its speed 100 km per hour from the usual speed find his usual speed

Answer 1

Hello Dear.

Let the time at which the Aeroplane originally reaches the destination be x hr.
Distance = 1500 km.
∴ Original speed of the Aeroplane = 1500/x
[∵ Speed = Distance/Time]


Now,
∵ Aeroplane starts after the 30 minutes (or 1/2 = 0.5 hrs.) then its scheduled time.
∴ Time taken by the Aeroplane to reach the destination = (x - 0.5) hrs.

∴ New Speed of the Aeroplane = 1500/(x - 0.5) 

Now,
According to the Question,
New Speed of the Aeroplane - Original Speed of the Aeroplane = 100 km/hr.
∴ 1500/(x - 0.5) - 1500/x = 100
⇒ 2x² - x - 15 = 0
∴ Splitting the middle term,
2x² - 6x + 5x - 15 = 0
⇒ 2x(x - 3) + 5(x - 3) = 0
⇒ (2x + 5)(x - 3) = 0
⇒ x = -5/2 and x = 3

∴ Time cannot be negative. ∴ Rejecting - 5/2 hrs.

Thus, time taken is 3 hrs.

Now, Usual Speed (or Original Speed) = 1500/3
 = 500 km/hr.


Hence, the usual speed of the Aeroplane is 500 km/hr.


Hope it helps.

Answer 2

Solution :-

Let the original speed of train be x km/hr

New speed = (x + 100) km/hr

We know that,

Time = Distance / Speed

Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.

According to the question,

=> 1500/x - 1500/(x + 100) = 1/2

=> (1500x + 15000 - 1500x)/x(x + 100) = 1/2

=> 2(15000) = x(x + 100)

=> 30000 = x² + 100x

=> x² + 100x - 30000 = 0

=> x² + 600x - 500x - 30000 = 0

=> x(x + 600) - 500(x + 600) = 0

=> (x - 500) (x + 600) = 0

=> x = 500 or x = - 600

∴ x ≠ - 600 (Because speed can't be negative)

Hence,

Its usual speed = 500 km/hr